If we integrate $f(z)=\dfrac{e^{iz}}{\sqrt{z}}$ around the contour $[0,R]\cup Re^{i[0,\pi/2]}\cup i[R,0]$ as $R\to\infty$, we get that
$$
\int_0^R\frac{e^{ix}}{\sqrt{x}}\,\mathrm{d}x
+\int_0^{\pi/2}\frac{e^{iRe^{ix}}}{\sqrt{R}e^{ix/2}}iRe^{ix}\,\mathrm{d}x
-\sqrt{i\,}\int_0^R\frac{e^{-x}}{\sqrt{x}}\,\mathrm{d}x
=0
$$
because there are no singularities of $f$ inside the contour. Then because
$$
\begin{align}
\left|\int_0^{\pi/2}\frac{e^{iRe^{ix}}}{\sqrt{R}e^{ix/2}}iRe^{ix}\,\mathrm{d}x\right|
&\le\sqrt{R}\int_0^{\pi/2}e^{-R\sin(x)}\,\mathrm{d}x\\
&\le\sqrt{R}\int_0^{\pi/2}e^{-2Rx/\pi}\,\mathrm{d}x\\
&\le\frac\pi{2\sqrt{R}}
\end{align}
$$
vanishes as $R\to\infty$, we have
$$
\int_0^\infty\frac{e^{ix}}{\sqrt{x}}\,\mathrm{d}x
=\sqrt{i\,}\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,\mathrm{d}x
$$

as claimed, where the blue-colored integration is taken along the curve starting from $+\infty$ to $2$, and then turning back to $+\infty$, which makes it cancel out. Therefore we obtain the desired result.

The problem in this calculation is that the starred equality is almost unable to be justified by any simple means.